Sketch the response of the following 2nd order system to the

Sketch the response of the following 2^nd order system to the initial conditions 10 d^2 x(t)/dt^2 + 5 dx(t)/dt + 200 x(t) = 0 IC x(0) = 1 x(0) = 0

Solution

Draw the response we have to solve this equation therefore we will make use of differential equation to solve this

We will represent d(x)/dt by m, therefore our equation becomes
       10m² +2.5 m + 200 = 0 this is an auiliary equation.

Now we solve it for the \"m\"
the value of m will be complex number
i,e - 0.125 + i 4.47 , -0.125 - i 4.47
    Now we will write the solution of the above equation that will be

X = e^pt (ACosq + B Sinq) where p = -0.125 is the real part of root and q = 4.47 is the imaginary part of root.

Now we have to put the initial conditions given i.e. X(0) = 1 , X\'(0) = 0
On putting these value we have the solution for constants A and B.
A = 1 , B = 0 .
Now we have the final solution as X(t) = e^{- 0.125 t} Cos 4.47t

Now we will put the value of t ans find the value of X and draw it on the graph the generated curve will our response curve.